
If a charged particle projected in a gravity-free room deflects, then
(A) There must be an electric field
(B) There must be a magnetic field
(C) Both fields cannot be zero
(D) None of these
Answer
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Hint :Using the formula for force being exerted on a particle because of magnetic field and electric field, we will be able to find the dependency of its intensity due to the components in it. This will help us to find the validity of the given options.
Force due to magnetic field: $ {F_m} = q(\vec V \times B) $
Where $ {F_m} $ is the force on particle due to magnetic field and is expressed in Newton $ (N) $ , $ q $ is the charge of the particle and is expressed in Coulombs $ (C) $ , $ \vec V $ is the velocity of the charged particle and is expressed in meter per second $ (m/s) $ and $ B $ is the intensity of the magnetic field and is expressed in Webers $ (W) $ .
Force due to electric field: $ {F_e} = q(\vec E) $
Where $ {F_e} $ is the force on particles due to electric fields and is expressed in Newton $ (N) $ and $ \vec E $ is the intensity of the electric field and is expressed in Amperes $ (A) $ .
Complete Step By Step Answer:
We know that the force exerted on a moving charged particle due to magnetic field is $ {F_m} = q(\vec V \times B) $ and due to electric field is $ {F_e} = q(\vec E) $ .
We can gather from above that their intensities depend on a lot of factors.
Now let’s solve this question option by option.
When a projected charged particle deflects in a gravity-free room, it will deflect because of the force field around it and due to its inherent velocity. Here, the electric and magnetic fields act on the moving particle but it does not solely contribute to the deflection. Velocity also plays an important role. The particle would have deflected even in the absence of any one of the two fields. This means one can be zero, but the other one must have a minimum intensity.
Therefore, options A, B and C are false.
In conclusion, the correct option is C.
Note :
Fields so make the charged particle deflect. If there was not any velocity in the particle, the presence of a field would be a must for the movement of it. But due to the presence of velocity, no field’s presence is a must.
Force due to magnetic field: $ {F_m} = q(\vec V \times B) $
Where $ {F_m} $ is the force on particle due to magnetic field and is expressed in Newton $ (N) $ , $ q $ is the charge of the particle and is expressed in Coulombs $ (C) $ , $ \vec V $ is the velocity of the charged particle and is expressed in meter per second $ (m/s) $ and $ B $ is the intensity of the magnetic field and is expressed in Webers $ (W) $ .
Force due to electric field: $ {F_e} = q(\vec E) $
Where $ {F_e} $ is the force on particles due to electric fields and is expressed in Newton $ (N) $ and $ \vec E $ is the intensity of the electric field and is expressed in Amperes $ (A) $ .
Complete Step By Step Answer:
We know that the force exerted on a moving charged particle due to magnetic field is $ {F_m} = q(\vec V \times B) $ and due to electric field is $ {F_e} = q(\vec E) $ .
We can gather from above that their intensities depend on a lot of factors.
Now let’s solve this question option by option.
When a projected charged particle deflects in a gravity-free room, it will deflect because of the force field around it and due to its inherent velocity. Here, the electric and magnetic fields act on the moving particle but it does not solely contribute to the deflection. Velocity also plays an important role. The particle would have deflected even in the absence of any one of the two fields. This means one can be zero, but the other one must have a minimum intensity.
Therefore, options A, B and C are false.
In conclusion, the correct option is C.
Note :
Fields so make the charged particle deflect. If there was not any velocity in the particle, the presence of a field would be a must for the movement of it. But due to the presence of velocity, no field’s presence is a must.
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